Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of `2^(@)` to the right with the vertical , the other pendulum makes an angle of `1^(@)` to the left of the vertical. What is the phase difference between the pendulums?

Options

A. `(pi)/(2)`
B. `(2)/(3)pi`
C. `(3)/(2)pi`
D. `pi`

Answer

B

Explanation

Let `theta_(1)` and `theta_(2)` be the angular displacement of first and second pendulums respectively at an instant. `theta_(0)` be the maximum angular displacement of each pendulum. Let `theta_(1)` and `theta_(2)` be the initial phases of the two pendulums. then `theta_(1)=theta_(0)sin(omegat+phi_(1))` . . . . (i)
and `theta_(2)=theta_(0) sin (omegat+phi_(2))` . . (ii)
For first pendulum, `theta_(1)=2^(2)=theta_(0)`
From (i), we get
`2=2sim(omegat+phi_(1))` or `sin(omegat+phi_(1))=1`
or `omegat+phi_(1)=90^(@)` . .. (iii)
For second pendulum, `theta_(2)=-1^(@)`
From (ii), we get
`-1=2sin(omegat+phi_(2))` or `sin(omegat+phi_(2))=-(1)/(2)`
or `sin(omegat+phi_(2))=sin(180^(@)+30^(@))=sin210^(@)`
`thereforeomegat+phi_(2)=210^(@)` . . . (iv)
`therefore`Subtracting (iii) from (iv), we get
`(omegat+phi_(2))-(omegat+phi_(1))=210^(@)-90^(@)=120^(@)=(2pi)/(3)`